A mass M, attached to a horizontal spring, ex | vedclass

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Question

The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.

$	herefore \quad M v_{1}=(M+m) v_{2}$

$M A_{1} \sqr

Vedclass · Accepted Answer

${\left( {\;\frac{{M + m}}{M}} \right)^{\frac{1}{2}}}$

Vedclass · Answer

The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.

$	herefore \quad M v_{1}=(M+m) v_{2}$

$M A_{1} \sqrt{\frac{k}{M}}=(M+m) A_{2} \sqrt{\frac{k}{m+M}} 	herefore \quad(V=A \sqrt{\frac{k}{M}})$

$A_{1} \sqrt{M}=A_{2} \sqrt{M+m} \quad 	herefore \frac{A_{1}}{A_{2}}=\sqrt{\frac{m+M}{M}}$

A mass $M$, attached to a horizontal spring, executes S.H.M. with amplitude $A_1$. When the mass $M$ passes through its mean position then a smaller mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\frac{{{A_1}}}{{{A_2}}}$ is

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